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1 change: 1 addition & 0 deletions properties/P000087.md
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Expand Up @@ -17,3 +17,4 @@ Contrary to Munkres or Willard, we do not assume any separation axiom like {P3},
#### Meta-properties

- This property is preserved by arbitrary products.
- This property is preserved by $\Sigma$-products.
1 change: 1 addition & 0 deletions properties/P000186.md
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Expand Up @@ -13,3 +13,4 @@ Homeomorphic to a subspace of a space that is {P87} and {P187}.

- This property is hereditary.
- This property is preserved by countable products.
- This property is preserved by $\Sigma$-products.
7 changes: 7 additions & 0 deletions spaces/S000035/properties/P000062.md
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---
space: S000035
property: P000062
value: false
---

Consider the open cover $\mathcal{O}=\{[0,a)\mid a \in X\}$. If $X$ were weakly Lindelöf, we would find a countable $\mathcal{U}\subseteq \mathcal{O}$ such that $\bigcup \mathcal{U}$ dense. But then $\bigcup\mathcal{U}=X$, since otherwise we would find an $a \in X$ with $[a,\omega_1)\subseteq X \setminus \bigcup\mathcal{U}$. Therefore $X$ would be a countable union of countable sets and thus countable.
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Consider the open cover $\mathcal{O}=\{[0,a)\mid a \in X\}$. If $X$ were weakly Lindelöf, we would find a countable $\mathcal{U}\subseteq \mathcal{O}$ such that $\bigcup \mathcal{U}$ dense. But then $\bigcup\mathcal{U}=X$, since otherwise we would find an $a \in X$ with $[a,\omega_1)\subseteq X \setminus \bigcup\mathcal{U}$. Therefore $X$ would be a countable union of countable sets and thus countable.
Consider the open cover $\mathcal{O}=\{[0,\alpha)\mid \alpha \in X\}$.
Every countable subset of $\omega_1$ has an upper bound in $\omega_1$.
So if $\mathcal U$ is a countable subcollection of $\mathcal O$, there is some $\alpha\in\omega_1$ such that
$\bigcup\mathcal U\subseteq[0,\alpha]$, which is not dense in $X$.

more direct

7 changes: 7 additions & 0 deletions spaces/S000035/properties/P000082.md
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---
space: S000035
property: P000082
value: true
---

For $a \in X$ consider the open neighborhood $U_a := [0,a+1)\subseteq X$. By the definition of $X$, $U_a$ is {P57} {P190} and thus {P53} [(Explore)](https://topology.pi-base.org/spaces?q=Countable+%2B+Ordinal+space+%2B+%7EMetrizable).
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For $a \in X$ consider the open neighborhood $U_a := [0,a+1)\subseteq X$. By the definition of $X$, $U_a$ is {P57} {P190} and thus {P53} [(Explore)](https://topology.pi-base.org/spaces?q=Countable+%2B+Ordinal+space+%2B+%7EMetrizable).
For $alpha \in X$ consider the open neighborhood $U_\alpha := [0,\alpha+1)\subseteq X$. By the definition of $X$, $U_\alpha$ is {P57} {P190} and thus {P53} [(Explore)](https://topology.pi-base.org/spaces?q=Countable+%2B+Ordinal+space+%2B+%7EMetrizable).

for ordinals it's more common to use Greek letters from the beginning of the alphabet